3 Answers
3

I don't know what you are asking but I will explain how I view this sort of thing.

What is meant by $$3x = \tan(\theta)$$ is that both sides are the same function of $x$ or $\theta$ and $x$ and $\theta$ are related somehow. Writing out the relation explicitly we have $$3x = \tan(\theta(x)).$$ Now define $f(x) = 3x$ and $g(x) = \tan(\theta(x))$, this equation means that $f = g$.

Applying the derivative operator to both sides $f' = g'$ we have $$3 = \theta'(x)\sec(\theta(x))^2$$ (by the chain rule and derivative of tan = sec^2). Now you can write it as a differential $$3 \mathrm{d}x = \sec(\theta(x))^2 \theta'(x) \mathrm{d}x = \sec(\theta)^2 \mathrm{d}\theta$$ since $\theta'(x) \mathrm{d}x = \mathrm{d}\theta$.

This is differential notation. If $y=f(x)$, then it is customary to write $dy=f'(x)dx$. What this says is that a small change in $x$, $dx$, produces an approximate change on $y$ of $f'(x)dx$. In your instance you have $x=f(\theta)$ and so $dx=f'(\theta)d\theta$. The concept comes into play when you talk about linear approximation and you are using the differentials to approximate error. So you have